2 Answers
2

T. Nagell in [Norsk Mat. Forenings Skrifter.1:4 (1921)]
shows that, for an odd prime $q$, the equation
$$
x^2-y^q=1 \qquad (*)
$$
has a solution in integers $x>1$, $y>1$, then $y$ is even and $q\mid x$.
In his proof of the latter divisibility he uses a similar trick as follows.

Assuming $q\nmid x$ write ($*$) as
$$
x^2=(y+1)\cdot\frac{y^q+1}{y+1}
$$
where the factors on the right are coprime (they could only have
common multiple $q$). Therefore,
$$
y+1=u^2, \quad \frac{y^q+1}{y+1}=v^2, \quad x=uv,
\qquad (u,v)=1, \quad \text{$u,v$ are odd}.
$$
Using these findings we can state the original equation in the form
$x^2-(u^2-1)^q=1$, or
$$
X^2-dZ^2=1 \quad\text{where $d=u^2-1$}.
$$
The latter equation has integral solution
$$
X=uv, \quad Z=(u^2-1)^{(q-1)/2},
$$
while its general solution (a classical result for this particular Pell's equation)
is taken the form $(u+\sqrt{u^2-1})^n$. It remains to use the binomial theorem in
$$
X+Z\sqrt{u^2-1}
=(u+\sqrt{u^2-1})^n
$$
(for certain $n\ge1$) and simple estimates to conclude that this is not possible.

To stress the use of similar trick: instead of showing insolvability of $x^2-(u^2-1)^q=1$,
we assume that a solution exists and then use $d=u^2-1$ to produce a solution $X,Z$ of $X^2-dZ^2=1$;
finally, the pair $X,Z$ cannot solve the resulting Pell's equation.
(Of course, it is hard to claim that this is exactly Trost's trick,
as here is a dummy variable but no discriminants, except the one for
Pell's equation. Trost's trick is less trickier to my taste. $\ddot\smile$)

Note that Nagell's result was crucial for showing that ($*$) does not have
integral solutions $x>1$, $y>1$ for a fixed prime $q>3$. This was shown in
a very elegant way, using the Euclidean algorithm and quadratic residues
by Ko Chao [Sci. Sinica14 (1965) 457--460], and later reproduced
in Mordell's Diophantine equations. The ideas of this proof are in the
heart of Mihailescu's ultimate solution of Catalan's conjecture. A much
simpler proof of Ko Chao's result, based on a completely different (nice!) trick,
was given later by E.Z. Chein [Proc. Amer. Math. Soc.56 (1976) 83--84].

Cassels reviews Trost's paper (MR0288077 (44 #5275)) and notes that 1) the equivalence of the unsolvability of $x^4-2y^4=z^2$ and $x^4+y^4=z^2$ is well-known (he cites Hardy and Wright, Theorem 226), and 2) "There is a second application of the same trick which shows that there are no non-trivial rational points on a certain elliptic curve if there are none on its jacobian."